Question

In Part A, we saw that ΔG∘=−242.1 kJ for the hydrogenation of acetylene under standard conditions (all pressures equal to 1 atm and the common reference temperature 298 K). In Part B, you will determine the ΔG for the reaction under a given set of nonstandard conditions.

Part B

At 25 ∘C the reaction from Part A has a composition as shown in the table below.

Substance	Pressure (atm)
C2H2(g)	3.95
H2(g)	5.65
C2H6(g)	5.25×10−2

What is the free energy change, ΔG, in kilojoules for the reaction under these conditions?

Answer 1

The reaction:

C₂H₂(g) + 2H₂(g) = C₂H₆(g)

Information:

• P C₂H₂(g) = 3.95atm
• P H₂(g) = 5.65 atm
• P C₂H₆(g) = 0.0525 atm
• R = 8.314 J/molK
• T = 298 K

1) Calculate Kp using the formula:

Kp = (P products) / (P reagents)

Use the balanced chemical equation and transform the equation into:

Kp = (P C₂H₆) / (P H₂)²(P C₂H₂)

Note: the superscript 2 in H₂ is there because in the balanced chemical equation, the coefficient of H₂ is 2.

Substitute the values:

Kp = (0.0525 atm) / (5.65 atm)²(3.95 atm)

Kp = 0.0004163 atm^-2

Calculate ΔG using the formula:

ΔG = -RTlnKp

Substitute the values:

ΔG = -(8.314 J/molK)(298K)ln(0.0004163)

ΔG = 19285.34 J/mol

ΔG = 19.3 kJ/mol (3 significant figures)

Answer: the reaction's ΔG is 19.3 kJ/mol.